Solve for $x$ and $y$ using substitution. ${x-6y = 3}$ ${x = 3y+6}$
Since $x$ has already been solved for, substitute $3y+6$ for $x$ in the first equation. ${(3y+6)}{- 6y = 3}$ Simplify and solve for $y$ $3y+6 - 6y = 3$ $-3y+6 = 3$ $-3y+6{-6} = 3{-6}$ $-3y = -3$ $\dfrac{-3y}{{-3}} = \dfrac{-3}{{-3}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = 3y+6}\thinspace$ to find $x$ ${x = 3}{(1)}{ + 6}$ $x = 3 + 6$ ${x = 9}$ You can also plug ${y = 1}$ into $\thinspace {x-6y = 3}\thinspace$ and get the same answer for $x$ : ${x - 6}{(1)}{= 3}$ ${x = 9}$